1、
由題意,得
(a1+2)/2=√(2a1)
整理,得
(a1-2)²=0
a1-2=0 a1=2
(an+2)/2=√(2Sn)
整理,得
8Sn=(an+2)²
8Sn-1=[a(n-1)+2]²
8an=(an+2)²-[a(n-1)+2]²=an²+4an-a(n-1)²-4a(n-1)
an²-a(n-1)²-4an-4a(n-1)=0
[an+(a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-4]=0
數(shù)列是正數(shù)數(shù)列,an+a(n-1)>0,要等式成立,只有an-a(n-1)-4=0
an-a(n-1)=4,為定值.
數(shù)列{an}是以2為首項(xiàng),4為公差的等差數(shù)列.
an=2+4(n-1)=4n-2
數(shù)列{an}的通項(xiàng)公式為an=4n-2
2、
bn=0.5[a(n+1)/an+an/a(n+1)]
=[(4n+2)/(4n-2)+(4n-2)/(4n+2)]/2
=[1+4/(4n-2)+1-4/(4n+2)]/2
=1+2/[(2n-1)(2n+1)]
=1+1/(2n-1)-1/(2n+1)
b1+b2+...+bn
=n+1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)
=n+1-1/(2n+1)